The average translational kinetic energy of N2 gas molecules at ...........°C becomes equal to the K

Q: Why do we use $\frac{3}{2} K_B T$ for $N_2$ when it is a diatomic gas? Shouldn't it be $\frac{5}{2} K_B T$?

The question specifically asks for the *translational* kinetic energy. The translational kinetic energy is always $\frac{3}{2} K_B T$ for any gas, regardless of whether it is monatomic, diatomic, or polyatomic. If the question had asked for the *total* kinetic energy, then you would use $\frac{5}{2} K_B T$ to account for the two rotational degrees of freedom.

Q: What is the difference between $K_B$ and $R$?

$K_B$ (Boltzmann constant) is the gas constant per *molecule*, while $R$ (Universal gas constant) is the gas constant per *mole*. They are related by the equation $R = N_A K_B$, where $N_A$ is Avogadro's number.

Q: Why is the kinetic energy of the electron exactly $eV$?

When a charge $q$ moves through a potential difference $V$, the work done by the electric field is $W = qV$. According to the work-energy theorem, if the particle starts from rest, all this work is converted into its kinetic energy. For an electron, $q = e$, so the kinetic energy is $eV$.

Q: Why did we get the temperature in Kelvin first?

The formula $\frac{3}{2} K_B T$ is derived from the ideal gas law and the kinetic theory of gases, which strictly require the use of absolute temperature measured in Kelvin. You must always calculate $T$ in Kelvin first and then convert it to Celsius if needed.

Q: What if the electron didn't start from rest?

If the electron had an initial velocity, the final kinetic energy would be the sum of its initial kinetic energy and the work done by the electric field: $KE_{\text{final}} = KE_{\text{initial}} + eV$. However, the problem explicitly states that it accelerates from rest.

Question

The average translational kinetic energy of $N_2$ gas molecules at ...........$^{\circ}	ext{C}$ becomes equal to the KE of an electron accelerated from rest through a potential difference of $0.1	ext{ V}$.
[Given, $K_B = 1.38 	imes 10^{-23}	ext{ J/K}$]

Sigma Percentile · Accepted Answer

The beauty of physics often lies in its ability to connect seemingly unrelated phenomena. In this problem, we are asked to bridge two completely different domains: the chaotic, microscopic world of thermodynamics and the precise, directed world of electrostatics.

Imagine two distinct physical setups. On one side, we have a container filled with nitrogen gas ($N_2$) molecules zipping around randomly. Their energy is purely a manifestation of the temperature of the gas. On the other side, we have a single electron, starting from rest, and accelerating through a potential difference of $0.1	ext{ V}$. The problem presents a fascinating scenario where the energy in both these cases is exactly the same!

### The Kinetic Theory Perspective

Let's first look at the gas. According to the Kinetic Theory of Gases, the average translational kinetic energy of *any* gas molecule, regardless of whether it is monatomic, diatomic, or polyatomic, depends only on its absolute temperature.

This relationship is given by the formula:
$$KE_{	ext{avg}} = \frac{3}{2}K_B T$$

Here, $K_B$ is the Boltzmann constant ($1.38 	imes 10^{-23}	ext{ J/K}$) and $T$ is the absolute temperature in Kelvin. It is crucial to note that the question specifically asks for the *translational* kinetic energy. If it had asked for the *total* kinetic energy of the diatomic $N_2$ gas, we would have to account for rotational degrees of freedom, making the formula $\frac{5}{2}K_B T$.

### The Electrostatics Perspective

Now, let's shift our focus to the electron. When a charged particle is accelerated through an electric potential difference, the work done on it by the electric field is converted entirely into its kinetic energy (assuming it starts from rest).

The kinetic energy gained by the electron is simply its charge multiplied by the potential difference:
$$KE_{e^-} = eV$$

Here, $e$ is the elementary charge ($1.6 	imes 10^{-19}	ext{ C}$) and $V$ is the potential difference ($0.1	ext{ V}$).

### Bridging the Worlds: The Master Equation

Since the problem states that these two energies are equal, we can set up our master equation by equating the two expressions:
$$\frac{3}{2}K_B T = eV$$

Now, we substitute the known values into this equation. We plug in the Boltzmann constant, the elementary charge of an electron, and the given potential difference:
$$\frac{3}{2} 	imes (1.38 	imes 10^{-23}) 	imes T = (1.6 	imes 10^{-19}) 	imes 0.1$$

### The Final Calculation and the Celsius Trap

It is time for some algebra. We need to isolate $T$. Be very careful with the powers of ten here, as it is a common place for silly mistakes.

$$T = \frac{2 	imes 1.6 	imes 10^{-19} 	imes 0.1}{3 	imes 1.38 	imes 10^{-23}}$$

$$T = \frac{0.32 	imes 10^{-19}}{4.14 	imes 10^{-23}}$$

Bringing the $10^{-23}$ to the numerator makes it $10^{23}$, giving us a net power of $10^4$:
$$T = \frac{0.32}{4.14} 	imes 10^4 \approx 773	ext{ K}$$

We have found the absolute temperature to be $773	ext{ K}$. However, there is a catch! The question specifically asks for the temperature in degrees Celsius ($^{\circ}	ext{C}$). Always read the requested units carefully before finalizing your answer.

To convert Kelvin to Celsius, we subtract $273$:
$$T(^{\circ}	ext{C}) = 773 - 273 = 500^{\circ}	ext{C}$$

And there we have it! The temperature at which the average translational kinetic energy of $N_2$ molecules equals the kinetic energy of an electron accelerated through $0.1	ext{ V}$ is exactly **$500^{\circ}	ext{C}$**.

Animated Solution for Physics - Thermodynamics: The average translational kinetic energy of $N_{2}$ gas molecules at ........... $^{\circ} C$ becomes equal to the KE of an electron accelerated from rest through a potential difference of $0.1 V$ . [Given, $K_{B} = 1.38 \times 1 0^{- 23} J/K$ ]

Enter Numerical Value:

Visualized Solution

\text{Physical Setup}

\text{Energy Equations}

\text{Equating and Substituting}

\text{Solving for Temperature } T

\text{Conversion to Celsius}

\text{Conceptual Extensions}

The Sigma Insight: Kinetic Theory of Gases

The Kinetic Theory Perspective

The Electrostatics Perspective

Bridging the Worlds

The Final Calculation and the Celsius Trap

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